Understanding Newton's Third Law

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Newton's Third Law states that for every action, there is an equal and opposite reaction, which is crucial for understanding momentum conservation. The discussion explores how this law applies in various scenarios, such as a person throwing a baseball in space or a blade breaking off a spinning fan. Participants clarify that while the action-reaction forces are equal and opposite, the overall momentum of the system must be conserved, meaning the larger mass will recoil when a smaller mass is released. The conversation emphasizes that the center of mass of the system must move in accordance with the momentum changes, regardless of whether the larger mass visibly slows down. Understanding these principles helps clarify misconceptions about forces and momentum in isolated systems.
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Homework Statement



I'm having trouble thinking about Newton's third law pairs and have been trying to find some help searching the Physics Forums archives.

Homework Equations



I have found alternative explanations for 3rd law pairs involving thrust in an escaped balloon or in a rocket - citing unequal internal forces instead of action-reaction:
https://www.physicsforums.com/showpost.php?p=433271&postcount=1"

But I can't think about this way about someone throwing a baseball in space, or tossing sandbags from a railroad car.One post that seems to describe my difficulty is https://www.physicsforums.com/showpost.php?p=2843399&postcount=6"

The Attempt at a Solution



I keep wanting to think that the reaction force has a "cause".
I know this is wrong thinking, so I am trying to find ways to understand the third law from the conservation of momentum.

Is this right? :

Suppose there is a fan whirling in space (somehow energy is supplied to keep the blades turning). Then a piece of one blade flies off the end and the fan's blades lose their source of energy to turn (at the same time). Will the action-reaction forces be the same as a person throwing a baseball in space and experiencing the reaction force of the thrown ball so that they are moved in a direction opposite to the ball? The momentum of the broken piece is equal and opposite to the rest of the fan, and the momentum of the person equal and opposite to the thrown ball? So I should think of a person throwing a ball as losing a certain amount of momentum when the ball is released? or gaining momentum in the opposite direction?

Thanks for helping.
 
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bunchie said:
Suppose there is a fan whirling in space (somehow energy is supplied to keep the blades turning).
Why does it need energy to keep the blades turning? What is slowing them down? Think of Newton's first law.
Then a piece of one blade flies off the end and the fan's blades lose their source of energy to turn (at the same time).
If it doesn't need energy to keep turning, why would the blades stop?
Will the action-reaction forces be the same as a person throwing a baseball in space and experiencing the reaction force of the thrown ball so that they are moved in a direction opposite to the ball? The momentum of the broken piece is equal and opposite to the rest of the fan, and the momentum of the person equal and opposite to the thrown ball? So I should think of a person throwing a ball as losing a certain amount of momentum when the ball is released? or gaining momentum in the opposite direction?
If a blade breaks from a spinning fan in space, is that fundamentally any different from a person throwing a baseball in space by rotating their arm and letting a ball go?

In both cases there has to be conservation of momentum. So the larger mass from which the smaller mass is launched must recoil with equal an opposite momentum. This does not mean that the rest of the fan slows its rotation, however. What it means is that the centre of mass of the rest of the fan has to move with momentum equal to and in a direction opposite to the momentum of the broken blade (as measured in the rest frame of the original fan).

AM
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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